Question:

Show that the energy required to build up the current \( I \) in a coil of inductance \( L \) is \( \frac{1}{2} L I^2 \). 

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The energy stored in an inductor is \( \frac{1}{2} L I^2 \), which represents the work done in establishing the magnetic field.
Updated On: Jun 13, 2025
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Solution and Explanation

Energy Stored in an Inductor 

Problem:

Show that the energy required to build up a current \( I \) in a coil of inductance \( L \) is:
\[ U = \frac{1}{2} L I^2 \]

Explanation and Derivation:

When a current flows through an inductor, it builds up a magnetic field. Energy is required to establish this magnetic field.

According to Faraday’s law of electromagnetic induction, the emf induced in the inductor is given by: \[ \mathcal{E} = L \frac{dI}{dt} \]

To build up the current from 0 to \( I \), work must be done against this self-induced emf. The small amount of work done in time \( dt \) is: \[ dW = \mathcal{E} \cdot I \cdot dt = L \frac{dI}{dt} \cdot I \cdot dt = L I \, dI \]

Total work done (energy stored) is: \[ U = \int_0^I L I \, dI = L \int_0^I I \, dI = L \left[ \frac{I^2}{2} \right]_0^I = \frac{1}{2} L I^2 \]

Conclusion:

Therefore, the energy stored in an inductor carrying current \( I \) is: \[ U = \frac{1}{2} L I^2 \]

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